Let \( i_{E} \), \( i_{C} \) and \( i_{B} \) represent the emitter current, the collector current and base current respectively in a transistor. Then:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( i_C \) is slightly smaller than \( i_E \)
By Kirchhoff’s current law applied to the transistor, the emitter current is the sum of the collector and base currents.
\( i_E = i_C + i_B \)
The base is thin and lightly doped, so only a small fraction of the emitter current leaves through the base, \( i_B \ll i_E \). Hence \( i_C = i_E - i_B \) is slightly smaller than \( i_E \), which is Option (1).
Option (2) would need a negative base current, and Options (3) and (4) make the small base current larger than the large emitter or collector current.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Transistor: Structure and Action – “… The current entering into the emitter from outside is equal to the emitter current \( I_E \). Similarly the current emerging from the base terminal is \( I_B \) and that from collector terminal is \( I_C \). It is obvious from the above description and also from a straight forward application of Kirchhoff’s law to Fig. 14.28(a) that the emitter current is the sum of collector current and base current:
\( \displaystyle I_E = I_C + I_B \qquad (14.7) \)
We also see that \( I_C \approx I_E \).”